Problem: Divide the following complex numbers. $ \dfrac{-13+i}{-1+2i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-1-2i}$ $ \dfrac{-13+i}{-1+2i} = \dfrac{-13+i}{-1+2i} \cdot \dfrac{{-1-2i}}{{-1-2i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-13+i) \cdot (-1-2i)} {(-1+2i) \cdot (-1-2i)} = \dfrac{(-13+i) \cdot (-1-2i)} {(-1)^2 - (2i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-13+i) \cdot (-1-2i)} {(-1)^2 - (2i)^2} = $ $ \dfrac{(-13+i) \cdot (-1-2i)} {1 + 4} = $ $ \dfrac{(-13+i) \cdot (-1-2i)} {5} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-13+i}) \cdot ({-1-2i})} {5} = $ $ \dfrac{{-13} \cdot {(-1)} + {1} \cdot {(-1) i} + {-13} \cdot {-2 i} + {1} \cdot {-2 i^2}} {5} $ Evaluate each product of two numbers. $ \dfrac{13 - 1i + 26i - 2 i^2} {5} $ Finally, simplify the fraction. $ \dfrac{13 - 1i + 26i + 2} {5} = \dfrac{15 + 25i} {5} = 3+5i $